How do you find the vertical, horizontal or slant asymptotes for y=(2x^2-3x+4)/(x+2)?

1 Answer
Jun 3, 2016

Vertical asymptotes at x_v = -2
Slant asymptote y = 2x-7

Explanation:

In a polynomial fraction f(x) = (p_n(x))/(p_m(x)) we have:

1) vertical asymptotes for x_v such that p_m(x_v)=0
2) horizontal asymptotes when n le m
3) slant asymptotes when n = m + 1
In the present case we have x_v = -2 and n = m+1 with n = 2 and m = 1

Slant asymptotes are obtained considering (p_n(x))/(p_{n-1}(x)) approx y = a+b x for large values of abs(x)

In the present case we have

(p_n(x))/(p_{n-1}(x)) = (2x^2-3x+4)/(x + 2)
p_n(x) = p_{n-1}(x)(a x + b) + r_{n-2}(x)
(p_n(x))/(p_{n-1}(x)) = (2x^2-3x+4) =(x+2) (a x +b)+c

equating we have

{ (4 - 2 b - c=0), (-3 - 2 a - b=0), (2 - a=0) :}

Solving for a,b,c we have a = 2,b=-7,c=18
so the slant asymptote reads

y = 2x-7

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